Two identical pendulums A and B are suspended | vedclass

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(A) Let the charges on the bobs be $q_A$ and $q_B$ such that $q_A > q_B$. Let the mass of each bob be $m$ and the length of the strings be $l$.In equi

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$	heta_1 = 	heta_2$

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(A) Let the charges on the bobs be $q_A$ and $q_B$ such that $q_A > q_B$. Let the mass of each bob be $m$ and the length of the strings be $l$.In equilibrium,the forces acting on each bob are: $1$. Gravitational force $mg$ acting downwards.$2$. Electrostatic repulsive force $F_e = \frac{k q_A q_B}{r^2}$ acting horizontally.$3$. Tension $T$ in the string.For each bob,the equilibrium condition is:$T \sin 	heta = F_e$$T \cos 	heta = mg$Dividing the two equations,we get:$	an 	heta = \frac{F_e}{mg}$Since the electrostatic force $F_e$ is the same for both bobs (by Newton's third law,the force exerted by $A$ on $B$ is equal to the force exerted by $B$ on $A$),and the masses $m$ are identical,the angles must be equal.Therefore,$	heta_1 = 	heta_2$.

Two identical pendulums $A$ and $B$ are suspended from the same point. The bobs are given positive charges,with $A$ having more charge than $B$. They diverge and reach equilibrium,with $A$ and $B$ making angles $\theta_1$ and $\theta_2$ with the vertical,respectively. Then:

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